A racket ball player hits a ball with a speed of 4.87 m/s the ball hits the court at a horizontal distanse of 1.95m at what hight did the ball leave the racket.?

Hi,
Question:-
A racket ball player hits a ball with a speed of 4.87 m/s the ball hits the court at a horizontal distanse of 1.95m at what hight did the ball leave the racket.?
Answer:-
1 - Horizontal speed is 4.87 m/s (assumption as there is ambiguity in the question about angle at start).
2 - distance travelled horizontally is 1.95 met.
3 - time of flight = 1.95 met / 4.87 met/sec = 0.4004 sec
4 - vertical downward speed at start = 0
5 - Vertical distance is s = 0.5*9.81*(0.4004*0.4004)=0.7864 met.

formula is s = ut + (1/2)gt^2

There is slight ambiguity in the question.
Science questions should be unambiguous.
The angle of the ball had to be assumed as 0 deg with horizontal axis.
It could be anything, and we could have found solution using cos and sin functions (components).

regards
Jayen